You are not logged in.
Offline
Offline
Offline
Offline
Offline
[align=center]
[/align]Offline
Offline
Offline
Offline
Yup!
Offline
Last edited by JaneFairfax (2009-03-22 03:33:42)
Offline
d=e=1 disproves both parts of #9.
(But not anymore.)
Why did the vector cross the road?
It wanted to be normal.
Offline
I misunderstood the term square free. I thought it was synonymous with non perfect square (and therefore excluded 1).
I have rephrased the question. It shouldnt be ambiguous now.
Offline
Offline
Offline
Offline
why has no one tried #1 ???
newayz ...
we can rewrite the equn as ... (n-2)(n-1)(n²-2)(n+1)(n+2)
we can clearly see that this expression will be div for n= 7k+1,7k+2,7k+5,7k+6 ... as in each of these casesone of n-1,n-2,n+1,n+2 ... contributes a factor of 7
for 7k+3 and 7k+4 we can find, by substitution, that n²-2 donates a factor of 7.
hence the expression is div by 7 ...
now, if n-1 and n+1 are even and as they are consecutive nos. we can write them as ... 2a,2a+2
now, 2a*(2a+2) = 4a*(a+1)
now a(a+1) will be div by 2! hence div of 8 is also true when n-1 is even
now if n-2 is even ... so is n²-2 and n+2 ....
3 even nos hence three multiples of 8 ... hence div by 8 ....
thus the exp is always div by 56 ...
I love Maths and Music ... dunno which more
Offline
@Kurre
Well done!
@smiyc86
Please use the hide tags.
Offline
Offline
@jane ... how to use the hidden text ??? some html codes reqd if so give the name of the starting tag ... btw i was going o use 7k-1 and 7k-2 instead of 7k+6 and 7k+5 .... but somehow didn't
I love Maths and Music ... dunno which more
Offline
I love Maths and Music ... dunno which more
Offline
Last edited by smiyc86 (2009-03-30 19:52:55)
I love Maths and Music ... dunno which more
Offline
Last edited by smiyc86 (2009-03-31 18:32:37)
I love Maths and Music ... dunno which more
Offline
smiyc86:
#6. Brilliant!
#9. Please expand on the proof.
Offline
@9 ... done
I love Maths and Music ... dunno which more
Offline